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I am to simplify $$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$$

into $\frac{2\sqrt{2x}+\sqrt{2}}{4}$

I am able to get to $\frac{x+4\sqrt{y}\sqrt{2}}{2}$ but cannot arrive at the provided solution.

Here is my working:

$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$ = $\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{64y}\sqrt{2}}$ = $\frac{x+4\sqrt{y}}{\sqrt{2}}$ = $\frac{x+4\sqrt{y}}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$ = $\frac{x+4\sqrt{y}\sqrt{2}}{2}$

I cannot see how to arrive at $\frac{2\sqrt{2x}+\sqrt{2}}{4}$?

Screen shot of my online textbooks question and answer in case I've typed it incorrectly: enter image description here

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  • $\begingroup$ Are you sure you copied the original equation and the answer correctly? I cannot see how you would have only one $x$ in the original expression, raised to the first power, which would somehow end up under the radical, i.e. $x^{1/2}$, in the answer. $\endgroup$ – Eevee Trainer Jan 9 at 1:16
  • $\begingroup$ In any event your own attempt is wrong. You cannot cancel the $\sqrt{64}$ like that because there wasn't a $\sqrt{64}$ coefficient for the $y$ term. $\endgroup$ – Eevee Trainer Jan 9 at 1:17
  • $\begingroup$ Have you tried factoring the $\sqrt{y}$ factor out? $\endgroup$ – ncmathsadist Jan 9 at 1:18
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    $\begingroup$ Okay, on seeing the edit, small note: $\sqrt{2} x$ means $\sqrt{2} \cdot x$, i.e. the $x$ isn't under the radical. I can see the confusion, I made the same mistake when I was younger. So before I even attempt to address the problem, I just wanted to note that. (If you want less ambiguity, $x \sqrt 2$ is also an acceptable way to write it.) $\endgroup$ – Eevee Trainer Jan 9 at 1:20
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    $\begingroup$ Yes, but the numerator is a sum. Split it up into its individual fractions, i.e. $$\frac{a+b}{c} = \frac a c + \frac b c $$ and you can see immediately why it doesn't follow. You would have to have the $\sqrt{64}$ in every term of the numerator and every term of the denominator for it to cancel. $\endgroup$ – Eevee Trainer Jan 9 at 1:22
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$$\frac{x \sqrt{64y} + 4 \sqrt{y}}{\sqrt{128 y}}$$ Factor our common factor $\sqrt{y}$ from numerator and denominator. $$\frac{\sqrt{y}(x \sqrt{64} + 4)}{\sqrt{y}(\sqrt{128})}$$ Notice $\sqrt{64} = \sqrt{8^2} = 8$ and $\sqrt{128} = \sqrt{64 * 2} = \sqrt{8^2 * 2} = 8\sqrt{2}$. $$\frac{8x + 4}{8\sqrt{2}}$$ Factor out $4$ from numerator and denominator and cancel. $$\frac{2x + 1}{2\sqrt{2}}$$ Multiply numerator and denominator by $\sqrt{2}$. $$\frac{2\sqrt{2}x + \sqrt{2}}{4}$$

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    $\begingroup$ It would be nice if you wrote more than just a few equations pulled out of thin air to explain your work. You know, explain what you did when and why. I feel that would be more helpful to the OP than just a bunch of equations with zero context. $\endgroup$ – Eevee Trainer Jan 9 at 1:23
  • $\begingroup$ Ok @EeveeTrainer, done. $\endgroup$ – Klint Qinami Jan 9 at 1:27
  • $\begingroup$ Thanks Klint. I can mostly follow. "Factor out 4 from numerator and denominator and cancel.", how do you do that? I cannot make the jump? $\endgroup$ – Doug Fir Jan 9 at 1:34
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    $\begingroup$ $\frac{8x + 4}{8\sqrt{2}} = \frac{4(2x + 1)}{4(2\sqrt{2})} = \frac{2x+1}{2 \sqrt{2}}$ $\endgroup$ – Klint Qinami Jan 9 at 1:35

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